What is the value of [ 1 - ( 90 - ) + ( 90 - ) ] [ ( 90 - ) - ( 90 - ) + 1 ]?

Answer: Option B

$$\begin{array}{l}\frac{\left[1-\tan \left(90-\theta \right)+\sec \left(90-\theta \right)\right]}{\left[\tan \left(90-\theta \right)-\sec \left(90-\theta \right)+1\right]}\\ \Rightarrow \frac{\left[1-\cot \theta +\mathrm{cosec}\theta \right]}{\left[\cot \theta +\mathrm{cosec}\theta +1\right]}\\ \Rightarrow \frac{\left[1-\frac{\cos \theta }{\sin \theta }+\frac{1}{\sin \theta }\right]}{\left[\frac{\cos \theta }{\sin \theta }+\frac{1}{\sin \theta }+1\right]}\\ \Rightarrow \frac{\left[\sin \theta -\cos \theta +1\right]}{\left[\sin \theta +\cos \theta +1\right]}\\ \Rightarrow \frac{\left(\sin \theta +1\right)-\cos \theta }{\left(\sin \theta +1\right)+\cos \theta }\\ \left[\begin{array}{l}\therefore \sin \theta =2\sin \frac{\theta }{2}.\cos \frac{\theta }{2}\\ \cos \theta =1-2{\sin }^2\frac{\theta }{2}\\ \cos \frac{\theta }{2}=2{\cos }^2\frac{\theta }{2}-1\end{array}\right]\\ \Rightarrow \frac{2\sin \frac{\theta }{2}.\cos \frac{\theta }{2}+1-1+2{\sin }^2\frac{\theta }{2}}{2\sin \frac{\theta }{2}.\cos \frac{\theta }{2}+1+2{\cos }^2\frac{\theta }{2}-1}\\ \Rightarrow \frac{2\sin \frac{\theta }{2}\left(\sin \frac{\theta }{2}+\cos \frac{\theta }{2}\right)}{2\cos \frac{\theta }{2}\left(\sin \frac{\theta }{2}+\cos \frac{\theta }{2}\right)}\\ \Rightarrow \tan \frac{\theta }{2}\end{array}$$